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Let p be a complex valued polynomial of two ral variables: $$ \sum {a_{ij} x^i y^j } $$ write: $$p(z)= \sum {P_j \overline z } ^j $$ where each $P_j$ is of the form $ P_j = \sum {b_{ij} z^i } $ Prove that p is an entire function if and only iff $ P_j \equiv 0 $

Clearly I have to consider the "derivate" $ P_{\overline z } = \frac{1} {2}\left( {P_x + iP_y } \right) $ , since the real and imaginary part are $C^1$ functions, being holomorphic it's equivalent of satisfy C.R , or equivalently $ P_{\overline z } = 0 $ in this case $$ P_{\overline z } = \sum {jP_j \overline z ^{j - 1} } = 0 $$ and now what can I do?

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Suppose some $P_j$ weren't 0 everywhere; how could you still get your last line? –  Kevin Carlson Aug 7 '12 at 0:29
    
intuitively I think that it's imposible to have that equality for all the "z" except in the case that I want to prove , but How can I prove it? –  Daniel Aug 7 '12 at 0:34
    
OK, as a start, suppose WLOG $P_1\neq 0$. Then $P_1\equiv -\sum_{j=2}^n jP_j \bar{z}^{j-1}$. So we've got a polynomial in $z$ equal to a polynomial in $z,\bar{z}$. Any thoughts on how we can rule this out? –  Kevin Carlson Aug 7 '12 at 0:47
    
@Kevin Carlson Maybe I'm bad , but thinking again, I can derivate a lot of times with respect to $ {\overline z } $ in the equality with zero, to deduce what I want to prove for example if $$ \eqalign{ & P\left( z \right) = P_0 \left( z \right) + P_1 \left( z \right)\overline z + P_2 \left( z \right)\overline z ^2 \cr & P_{\overline z } = P_1 \left( z \right) + 2P_2 \left( z \right)\overline z = 0 \cr & \left( {P_{\overline z } } \right)_{\overline z } = 0 = 2P_2 \left( z \right) \cr} $$ and using an inductive argument I have what I want –  Daniel Aug 7 '12 at 1:12
    
Hmm...the issue there is that a polynomial in $z$ doesn't have derivative 0 relative to $\bar{z}$: rather, that derivative doesn't exist. This is what I was going for before-that the left-hand side I'd gotten to was complex differentiable, while the right-hand side wasn't, unless both were constant. –  Kevin Carlson Aug 7 '12 at 1:18

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Hey Daniel I think your inductive argument makes sense-this is me just writing it out to make sure I understand, more or less :p. Namely, induct on the highest power of $j$ present in $P$. The claim is that $P_{\bar{z}} = 0 \Rightarrow P_j = 0 \in \mathbb{C}[z]$ for $j \geqslant 1$.

For $j = 1$, our equation $P_{\bar{z}}$ reads $P_1 = 0$ as a function, so this is the statement that if $p \in \mathbb{C}[z]$ evaluates to 0 everywhere, then $p$ is the 0 polynomial. Indeed, were it of degree $n$, for $n > 0$, it could have at most $n$ distinct roots, and $\mathbb{C}$ is infinite.

Inductively, to see that $j' < j \Rightarrow j$, suppose we have $$\frac{\partial}{\partial \bar{z}} \sum_{i \leqslant j} P_i \bar{z}^i = \sum_{i \leqslant j} i P_i \bar{z}^{i-1} = 0 \quad \in Fun(\mathbb{C}, \mathbb{C})$$Pulling the highest power to the other side, we have $$-jP_{j} \bar{z}^{j - 1} = \sum_{i < j} i P_i \bar{z}^{i - 1}$$Now apply $\frac{\partial^{j - 1}}{\partial \bar{z}^{j - 1}} \; \; \; $ to both sides, we get (up to some factorials $C$) that $$C P_j = 0 \Rightarrow P_j = 0$$by our inductive hypothesis we deduce that $P_i = 0$ for $i < j$ as well.

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